Question: Let $\mathbf{a} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}.$  Find the vector $\mathbf{b}$ such that $\mathbf{a} \cdot \mathbf{b} = 11$ and
\[\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -13 \\ -9 \\ 7 \end{pmatrix}.\]
Solution: Let $\mathbf{b} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}.$  Then the equation $\mathbf{a} \cdot \mathbf{b} = 11$ gives us $2x + y + 5z = 11.$  Also,
\[\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -5y + z \\ 5x - 2z \\ -x + 2y \end{pmatrix}.\]Comparing entries, we obtain
\begin{align*}
-5y + z &= -13, \\
5x - 2z &= -9, \\
-x + 2y &= 7.
\end{align*}Solving this system, along with the equation $2x + y + z = 5z = 11,$ we find $x = -1,$ $y = 3,$ and $z = 2.$  Hence, $\mathbf{b} = \boxed{\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}}.$